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1

I think $$f(x)=\cfrac{x}{1+\cfrac{x}{1+\cfrac{x}{1+\ddots}}}$$ isn't automatically well defined - it isn't clear what it is a limit of. To try and make this more precise, I've considered, for $x \in \mathbb{C}\setminus \left\{0\right\}$ and $z \in \widehat{\mathbb{C}}$ $$f(x,z)=\lim_{n\rightarrow \infty} g_{x}(z).$$ Where ...

0

From the equation $$f(x) = \frac{x}{1 + f(x)}$$ we get $$f(x) = \frac{1}{2}\left(\pm \sqrt{1+4 x} -1 \right)$$ Obviously the $+$ solution is the right one. Expanding the square root in a series we get: $$f(x) = C^{\frac{1}{2}}_1 2 x + C^{\frac{1}{2}}_2 8 x^2 + O(x^3)$$ ...

9

From the definition, you get $$\frac{x}{1+f(x)}=f(x)$$ Thus $$x=(1+f(x))f(x)$$ Now differentiate $$1=f'(x)+2f(x)f'(x)=f'(x)(1+2f(x))$$ Or $$f'(x)=\frac{1}{1+2f(x)}$$ Now, $f(0)=0$, from definition of $f$. One remark though: I did not prove the continued fraction is differentiable or even convergent. I just assume it's true, then you can compute ...

0

We want to apply the generating-function method with Hayman's method to determine asymptotics of the coefficients, as in LINK. Let \begin{align*} p_n &= (6n^2+6n+1) p_{n-1}+(4n^2-9n^4) p_{n-2}, \quad p_{-1} = 1, \quad p_0=1, \\ q_n &= (6n^2+6n+1) q_{n-1}+(4n^2-9n^4)q_{n-2}, \quad q_{-1} = 0, \quad q_0=1, \\ r_n &= \frac{p_n}{q_n}. ... 0 Let x = [a_0; a_1 ,a_2 , \ldots]. Find the continued fraction using the common algorithm. You will get x = [2;\overline{1,2}]. Sidenote: If you are confused about the inequalities a_n/b_n < x < c_n/d_n, remember that C_{2k} < C_{2k+1}, and C_{2k} < C_{2(k+1)} so that C_0 < C_2 < \ldots < C_n < C_{n-1} < \ldots < C_5 ... 1 You can find the proof in the book Khinchin, Continued fractions 0 Ok, So I found a simple inductive proof which relies on strengthening the inductive hypothesis (and hence the result). Proof that \forall i, d_i\in\mathbb{Z} and d_i|n-x_i^2 Base case: d_0=1 so it's \in\mathbb{Z} and it divides any integer expression Inductive case: Given some i, assume d_i\in\mathbb{Z} and d_i|n-x_i^2 Now because a_i ... 0 I don't understand. The continued fraction for \sqrt {41} is \langle 6; 2,2,12 \rangle. Here is output from my C++ program doing Lagrange's method of "reduced forms," which are the triples of integers in the middle of each line in the cycle. It is in precisely this situation, by the way, that Lagrange's method gives double the cyclic part in the ... 1 You already have:\frac{p_i}{p_{i-1}}=a_i + \frac1{\dfrac{p_{i-1}}{p_{i-2}}}$$Which is the induction step, in a proof by induction. Complete procedure: Let's prove by induction that, for i\ge1, then \dfrac{p_i}{p_{i-1}}=[a_i,a_{i-1},\dots,a_0]. The analogous proof for q_i is left to the reader. Induction base: First we test the induction base. ... 5 Because people commonly use the word evaluate in that context; this is simply a fact of usage. You can see other examples here on the website of the National Institute of Standards and Technology, in the title of this paper in SIAM Review, and here, to pick three of the first few examples that turned up on a search. In this context evaluate simply isn’t the ... 0 The convergents of a continued fraction alternate above and below the final value. There are more formal ways of showing this, but for now simply consider that a continued fraction can be formed by truncating the integer part and taking the reciprocal of the remainder at each step. Since truncating always takes a lower value, it is apparent that the first ... 0 Certainly C_0 < x, since \displaystyle x = a_0 + \frac{1}{[a_1;a_2,a_3\cdots]} and the fraction is positive. On the other hand, \displaystyle \frac{1}{a_1 + \frac{1}{[a_2;a_3,a_4\cdots]}} < \frac{1}{a_1} since the fraction in the denominator is positive. Thus \displaystyle x = a_0 + \frac{1}{[a_1;a_2,a_3\cdots]} = a_0 + \frac{1}{a_1 + ... 1 This follows from standard properties of convergents/continued fractions, and a fair bit of induction sprinkled throughout. We start by proving some standard properties of convergents/continued fractions; you may wish to skip straight to the main proof below. Lemma 1:$$\xi=<a_0, a_1, \ldots , a_{n-1}, \xi_n> Proof: We proceed by induction. When ...

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